Question

How do you integrate $\int{{{\sin }^{4}}\left( 2x \right)dx}$?

Answer 1

Hint: We first try to convert the trigonometric form into its multiple angle form. We use the theorem of $\cos 2\alpha =1-2{{\sin }^{2}}\alpha =2{{\cos }^{2}}\alpha -1$ and replace the square terms. Then to integrate the ratio cos we use the theorem $\int{\cos mxdx}=\dfrac{\sin mx}{m}+c$ and find the final solution.

Complete step by step solution:
We first use the theorem of multiple angles to find the simplified form of ${{\sin }^{4}}\left( 2x \right)$.
We know that $\cos 2\alpha =1-2{{\sin }^{2}}\alpha =2{{\cos }^{2}}\alpha -1$ which gives $2{{\sin }^{2}}\alpha =1-\cos 2\alpha $.
We replace the value of $\alpha $ with $2x$ and reform ${{\sin }^{4}}\left( 2x \right)$ as
${{\sin }^{4}}\left( 2x \right)={{\left( {{\sin }^{2}}2x \right)}^{2}}=\dfrac{{{\left( 2{{\sin }^{2}}2x \right)}^{2}}}{4}$.
Now we replace the values to get $\dfrac{{{\left( 2{{\sin }^{2}}2x \right)}^{2}}}{4}=\dfrac{{{\left( 1-\cos 4x \right)}^{2}}}{4}$.
Now we break the square part to get ${{\left( 1-\cos 4x \right)}^{2}}=1+{{\cos }^{2}}4x-2\cos 4x$.
We again the use the theorem of multiple angles $2{{\cos }^{2}}\alpha =1+\cos 2\alpha $ for ${{\cos }^{2}}4x$ and get
\[{{\cos }^{2}}4x=\dfrac{2{{\cos }^{2}}4x}{2}=\dfrac{1+\cos 8x}{2}\].
So, the final form of the replacement is
\[{{\sin }^{4}}\left( 2x \right)=\dfrac{{{\left( 1-\cos 4x \right)}^{2}}}{4}=\dfrac{1+{{\cos }^{2}}4x-2\cos 4x}{4}=\dfrac{1+\dfrac{1+\cos 8x}{2}-2\cos 4x}{4}\].
We simplify the equation to get \[{{\sin }^{4}}\left( 2x \right)=\dfrac{3}{8}+\dfrac{\cos 8x}{8}-\dfrac{\cos 4x}{2}\].
We now replace the value in the integration to get
$\int{{{\sin }^{4}}\left( 2x \right)dx}=\int{\left[ \dfrac{3}{8}+\dfrac{\cos 8x}{8}-\dfrac{\cos 4x}{2} \right]dx}$.
We know that $\int{\cos mx dx}=\dfrac{\sin mx}{m}+c$. We apply the rule to get
$\begin{align}
  & \int{\left[ \dfrac{3}{8}+\dfrac{\cos 8x}{8}-\dfrac{\cos 4x}{2} \right]dx} \\
 & =\dfrac{3}{8}x+\dfrac{\sin 8x}{64}-\dfrac{\sin 4x}{8}+c \\
\end{align}$
Therefore, the integral form of $\int{{{\sin }^{4}}\left( 2x \right)dx}$ is $\dfrac{3}{8}x+\dfrac{\sin 8x}{64}-\dfrac{\sin 4x}{8}+c$.

Note: In any cases of trigonometric multiple or submultiple angles, the main aim to solve the problem is to convert the square terms into multiple angle form without any squares. In that cases the integral form of we only have to find the forms for $\int{\cos mx dx}=\dfrac{\sin mx}{m}+c$ or $\int{\sin mxdx}=\dfrac{-\cos mx}{m}+c$.